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For $E:y^2=x^3+a{x}^2+bx$ with rational 2-torsion $(0,0)$, the quotient by that point is the two-isogenous curveThe two-isogeny descent maps a nonexceptional point to the square class of its $x$-coordinate, with $(0,0)$ mapping to $[b]$. The images are finite collections of squarefree divisors of $b$ and $a^2-4b$, determined by testing the associated homogeneous quartics for rational points. If their orders are $2^s$ and $2^{s^{\prime }}$, thenwhich determines the Mordell-Weil rank $r$.

The method requires a rational 2-isogeny, and deciding whether every locally soluble quartic is globally soluble can be difficult. Computing only local conditions gives a 2-isogeny Selmer group and hence an upper bound; a nontrivial Tate-Shafarevich group can make that bound strict. Even after finding the rank, a separate saturation and point search may be needed to find generators.

Solved by gpt-5.6-sol high.

Because the canonical height of an elliptic curve is a quadratic form, polarization makesa symmetric bilinear form on the free part of the Mordell-Weil group. If $P_i^{\prime }={\sum }_j{U}_{ij}{P}_j$ is another integral basis, then $U\in {\mathrm{GL}}_r(\mathbb{Z})$ and the Gram matrices satisfySince $\det U=\pm 1$, their determinants agree. Thus the regulator of an elliptic curve is independent of the chosen basis.

Now let $Q_1,\dots ,Q_r$ be a basis for the free part of $E^{\prime }(\mathbb{Q})$. The images $\varphi (P_i)$ span a finite-index sublattice, so modulo torsionfor an integral matrix $A$ with nonzero determinant. The height identity givesTaking determinants in the two descriptions of this Gram matrix yieldsTherefore the required formula holds with $d=(\det A{)}^{-1}\in {\mathbb{Q}}^{\times }$.

Solved by gpt-5.6-sol high.